Find the winner of the game to build the lexicographically smaller string
Two players are playing a game where string str is given. The first player can take the characters at even indices and the second player can take the characters at odd indices. The player which can build the lexicographically smaller string than the other player wins the game. Print the winner of the game, either player A, B or print Tie if it’s a tie.
Examples:
Input: str = “w3wiki”
Output: B
Explanation: “eeggoss” is the lexicographically smallest
string that player A can get.
“eefkkr” is the lexicographically smallest
string that player B can get.
And B’s string is lexicographically smaller.Input: str = “abcdbh”
Output: A
Approach: Create two empty strings str1 and str2 for players A and B respectively. Traverse the original string character by character and for every character whose index is even, append this character in str1 else append this character in str2. Finally, sort the generated string in order to get the lexicographically smallest possible string and compare them to find the winner of the game.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to find the winner of the game void find_winner(string str, int n) { // To store the strings for both the players string str1 = "" , str2 = "" ; for ( int i = 0; i < n; i++) { // If the index is even if (i % 2 == 0) { // Append the current character // to player A's string str1 += str[i]; } // If the index is odd else { // Append the current character // to player B's string str2 += str[i]; } } // Sort both the strings to get // the lexicographically smallest // string possible sort(str1.begin(), str1.end()); sort(str2.begin(), str2.end()); // Compare both the strings to // find the winner of the game if (str1 < str2) cout << "A" ; else if (str2 < str1) cout << "B" ; else cout << "Tie" ; } // Driver code int main() { string str = "w3wiki" ; int n = str.length(); find_winner(str, n); return 0; } |
Java
// Java implementation of the approach import java.util.Arrays; class GFG { // Function to find the winner of the game static void find_winner(String str, int n) { // To store the strings for both the players String str1 = "" , str2 = "" ; for ( int i = 0 ; i < n; i++) { // If the index is even if (i % 2 == 0 ) { // Append the current character // to player A's string str1 += str.charAt(i); } // If the index is odd else { // Append the current character // to player B's string str2 += str.charAt(i); } } // Sort both the strings to get // the lexicographically smallest // string possible char a[] = str1.toCharArray(); Arrays.sort(a); char b[] = str2.toCharArray(); Arrays.sort(b); str1 = new String(a); str2 = new String(b); // Compare both the strings to // find the winner of the game if (str1.compareTo(str2) < 0 ) System.out.print( "A" ); else if (str1.compareTo(str2) > 0 ) System.out.print( "B" ); else System.out.print( "Tie" ); } // Driver code public static void main(String[] args) { String str = "w3wiki" ; int n = str.length(); find_winner(str, n); } } // This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach # Function to find the winner of the game def find_winner(string, n): # To store the strings for both the players string1 = "" string2 = "" for i in range (n): # If the index is even if (i % 2 = = 0 ): # Append the current character # to player A's string string1 + = string[i] # If the index is odd else : # Append the current character # to player B's string string2 + = string[i] # Sort both the strings to get # the lexicographically smallest # string possible string1 = "".join( sorted (string1)) string2 = "".join( sorted (string2)) # Compare both the strings to # find the winner of the game if (string1 < string2): print ( "A" , end = "") elif (string2 < string1): print ( "B" , end = "") else : print ( "Tie" , end = "") # Driver code if __name__ = = "__main__" : string = "w3wiki" n = len (string) find_winner(string, n) # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approach using System; class GFG { // Function to find the winner of the game static void find_winner(String str, int n) { // To store the strings for both the players String str1 = "" , str2 = "" ; for ( int i = 0; i < n; i++) { // If the index is even if (i % 2 == 0) { // Append the current character // to player A's string str1 += str[i]; } // If the index is odd else { // Append the current character // to player B's string str2 += str[i]; } } // Sort both the strings to get // the lexicographically smallest // string possible char [] a = str1.ToCharArray(); Array.Sort(a); char [] b = str2.ToCharArray(); Array.Sort(b); str1 = new String(a); str2 = new String(b); // Compare both the strings to // find the winner of the game if (str1.CompareTo(str2) < 0) Console.Write( "A" ); else if (str1.CompareTo(str2) > 0) Console.Write( "B" ); else Console.Write( "Tie" ); } // Driver code public static void Main(String[] args) { String str = "w3wiki" ; int n = str.Length; find_winner(str, n); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript implementation of the approach // Function to find the winner of the game function find_winner(str, n) { // To store the strings for both the players var str1 = "" , str2 = "" ; for ( var i = 0; i < n; i++) { // If the index is even if (i % 2 == 0) { // Append the current character // to player A's string str1 += str[i]; } // If the index is odd else { // Append the current character // to player B's string str2 += str[i]; } } // Sort both the strings to get // the lexicographically smallest // string possible str1 = str1.split( '' ).sort(); str2 = str2.split( '' ).sort(); // Compare both the strings to // find the winner of the game if (str1 < str2) document.write( "A" ); else if (str2 < str1) document.write( "B" ); else document.write( "Tie" ); } // Driver code var str = "w3wiki" ; var n = str.length; find_winner(str, n); </script> |
B
Time Complexity: O(n log(n)), Where n is the length of the given string.
Auxiliary Space: O(n), for storing the strings for both the players.